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Question Number 110948 by bemath last updated on 01/Sep/20
     (√(bemath))  If each point on the line 3x+4y=2  is transformed by matrix M= (((2   0)),((0   1)) ) , the  image is a line ___
$$\:\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\mathrm{If}\:\mathrm{each}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}+\mathrm{4y}=\mathrm{2} \\ $$$$\mathrm{is}\:\mathrm{transformed}\:\mathrm{by}\:\mathrm{matrix}\:\mathrm{M}=\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:,\:\mathrm{the} \\ $$$$\mathrm{image}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\_\_\_ \\ $$
Commented by bemath last updated on 01/Sep/20
hshahaha....bemath sir. bemath macho
$$\mathrm{hshahaha}….\mathrm{bemath}\:\mathrm{sir}.\:\mathrm{bemath}\:\mathrm{macho} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/20
Thank you my dear!
$$\mathcal{T}{hank}\:{you}\:{my}\:{dear}! \\ $$
Commented by bobhans last updated on 01/Sep/20
standart way    (((x′)),((y′)) ) =  (((2    0)),((0    1)) )  ((x),(y) ) ⇒  ((x),(y) ) = (1/2) (((1   0)),((0   2)) )  (((x′)),((y′)) )   ((x),(y) ) = (1/2) (((x′)),((2y′)) ) =  ((((1/2)x′)),((   y′)) )  hence the image of line is (3/2)x′+4y′=2  multiply by 2 ⇒ 3x+8y−4=0
$$\mathrm{standart}\:\mathrm{way}\: \\ $$$$\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:\Rightarrow\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{2y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}'}\\{\:\:\:\mathrm{y}'}\end{pmatrix} \\ $$$$\mathrm{hence}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{line}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}'+\mathrm{4y}'=\mathrm{2} \\ $$$$\mathrm{multiply}\:\mathrm{by}\:\mathrm{2}\:\Rightarrow\:\mathrm{3x}+\mathrm{8y}−\mathrm{4}=\mathrm{0} \\ $$
Commented by Rasheed.Sindhi last updated on 01/Sep/20
 (√(bemath))=(√b)(√e)(√m)(√a)(√t)(√h)   :)
$$\left.\:\sqrt{\mathrm{bemath}}=\sqrt{\mathrm{b}}\sqrt{\mathrm{e}}\sqrt{\mathrm{m}}\sqrt{\mathrm{a}}\sqrt{\mathrm{t}}\sqrt{\mathrm{h}}\:\:\::\right) \\ $$
Commented by bemath last updated on 01/Sep/20
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Commented by Rasheed.Sindhi last updated on 01/Sep/20
BTW, be-math or bem-ath  or something else?
$$\mathrm{BTW},\:{be}-{math}\:{or}\:{bem}-{ath} \\ $$$${or}\:{something}\:{else}? \\ $$
Answered by bobhans last updated on 01/Sep/20
say the image is a line ℓ≡ x determinant (((3   4)),((0   1)))+y determinant (((2   0)),((3   4)))−2. determinant (((2    0)),((0    1)))=0  ℓ≡ 3x +8y−4 = 0
$$\mathrm{say}\:\mathrm{the}\:\mathrm{image}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\ell\equiv\:\mathrm{x}\begin{vmatrix}{\mathrm{3}\:\:\:\mathrm{4}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{vmatrix}+\mathrm{y}\begin{vmatrix}{\mathrm{2}\:\:\:\mathrm{0}}\\{\mathrm{3}\:\:\:\mathrm{4}}\end{vmatrix}−\mathrm{2}.\begin{vmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\ell\equiv\:\mathrm{3x}\:+\mathrm{8y}−\mathrm{4}\:=\:\mathrm{0}\: \\ $$

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